Certification Problem
Input (TPDB TRS_Standard/SK90/2.26)
The rewrite relation of the following TRS is considered.
|
f(0) |
→ |
0 |
(1) |
|
f(s(0)) |
→ |
s(0) |
(2) |
|
f(s(s(x))) |
→ |
p(h(g(x))) |
(3) |
|
g(0) |
→ |
pair(s(0),s(0)) |
(4) |
|
g(s(x)) |
→ |
h(g(x)) |
(5) |
|
h(x) |
→ |
pair(+(p(x),q(x)),p(x)) |
(6) |
|
p(pair(x,y)) |
→ |
x |
(7) |
|
q(pair(x,y)) |
→ |
y |
(8) |
|
+(x,0) |
→ |
x |
(9) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(10) |
|
f(s(s(x))) |
→ |
+(p(g(x)),q(g(x))) |
(11) |
|
g(s(x)) |
→ |
pair(+(p(g(x)),q(g(x))),p(g(x))) |
(12) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
6 |
|
stat(f) |
= |
lex
|
| prec(0) |
= |
2 |
|
stat(0) |
= |
mul
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
mul
|
| prec(h) |
= |
4 |
|
stat(h) |
= |
mul
|
| prec(g) |
= |
5 |
|
stat(g) |
= |
mul
|
| prec(pair) |
= |
0 |
|
stat(pair) |
= |
lex
|
| prec(+) |
= |
3 |
|
stat(+) |
= |
mul
|
| π(f) |
= |
[1] |
| π(0) |
= |
[] |
| π(s) |
= |
[1] |
| π(p) |
= |
1 |
| π(h) |
= |
[1] |
| π(g) |
= |
[1] |
| π(pair) |
= |
[2,1] |
| π(+) |
= |
[1,2] |
| π(q) |
= |
1 |
all of the following rules can be deleted.
|
f(0) |
→ |
0 |
(1) |
|
f(s(0)) |
→ |
s(0) |
(2) |
|
f(s(s(x))) |
→ |
p(h(g(x))) |
(3) |
|
g(0) |
→ |
pair(s(0),s(0)) |
(4) |
|
g(s(x)) |
→ |
h(g(x)) |
(5) |
|
h(x) |
→ |
pair(+(p(x),q(x)),p(x)) |
(6) |
|
p(pair(x,y)) |
→ |
x |
(7) |
|
q(pair(x,y)) |
→ |
y |
(8) |
|
+(x,0) |
→ |
x |
(9) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(10) |
|
f(s(s(x))) |
→ |
+(p(g(x)),q(g(x))) |
(11) |
|
g(s(x)) |
→ |
pair(+(p(g(x)),q(g(x))),p(g(x))) |
(12) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.